\newproblem{lay:6_3_24}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.3.24}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $W$ be a subspace of $\mathbb{R}^n$ with an orthogonal basis $\{\mathbf{w}_1,\mathbf{w}_2,...,\mathbf{w}_p\}$, and let
	$\{\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_q\}$ be an ortohogonal basis for $W^\perp$.
	\begin{enumerate}[a.]
		\item Explain why $\{\mathbf{w}_1,\mathbf{w}_2,...,\mathbf{w}_p,\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_q\}$ is an orthogonal set.
		\item Explain why the set in part (a) spans $\mathbb{R}^n$.
		\item Show that $\dim\{W\}+\dim\{W^\perp\}=n$
	\end{enumerate}
}{
   % Solution
	\begin{enumerate}[a.]
		\item Since both sets $\{\mathbf{w}_1,\mathbf{w}_2,...,\mathbf{w}_p\}$ and $\{\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_q\}$
		      are orthogonal bases, all products $\mathbf{w}_i\cdot\mathbf{w}_j=0=\mathbf{v}_i\cdot\mathbf{v}_j$ (for $i\neq j$).
					We still need to show that the products $\mathbf{w}_i\cdot\mathbf{v}_j=0$ for any $i\in\{1,2,...,p\}$ and $j\in\{1,2,...,q\}$. But this
					is true since $\mathbf{w}_i\in W$ and $\mathbf{v}_j\in W^\perp$.
		\item Since $W$ and $W^\perp$ are perpendicular spaces contained in $\mathbb{R}^n$ by the Orthogonal Decomposition Theorem (Theorem 6.3.8) we have
		      that any vector can be decomposed as a sum of a vector in $W$ and a vector in $W^\perp$. But any vector in $W$ can be expressed as a linear combination
					of the $\mathbf{w}_i$ vectors and any vector in $W^\perp$ can be expressed as a linear combination
					of the $\mathbf{v}_i$ vectors. So the combined set $\{\mathbf{w}_1,\mathbf{w}_2,...,\mathbf{w}_p,\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_q\}$ can generate
					both parts of the orthogonal decomposition, and consequently, can generate any vector in $\mathbb{R}^n$. In fact, since all the vectors in the set
					are orthogonal, the set is a basis of $\mathbb{R}^n$.
		\item We know that $\dim\{W\}=p$ and $\dim\{W^\perp\}=q$. We need to show that $p+q=n$. But this is true since the set in part (a) has $p+q$ vectors, and
				  we have stated in part (b) that these $p+q$ vectors is a basis for $\mathbb{R}^n$, so it must have exactly $n$ vectors.
	\end{enumerate}
}
\useproblem{lay:6_3_24}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
